swift 3 : Encode String to UTF-16LE

Question

我需要将字符串编码为 UTF-16LE(稍后转换为 sha1)，但我遇到了一些问题。这是我试过的:

let utf16array = Array("password".utf16)
print(utf16array)
// [112, 97, 115, 115, 119, 111, 114, 100]

但这正是我所期待的:

// [112, 0, 97, 0, 115, 0, 115, 0, 119, 0, 111, 0, 114, 0, 100, 0] 

同样的事情使用 utf8array:

let utf8array = "password".utf8.map({ $0 as UInt8 })
// [112, 97, 115, 115, 119, 111, 114, 100]

所以，这就是我“修复”它所做的:

var bytesArray:[UInt16] = []
for byte in utf16array {
    bytesArray.append(byte)
    bytesArray.append(0)
}
print(bytesArray)
// [112, 0, 97, 0, 115, 0, 115, 0, 119, 0, 111, 0, 114, 0, 100, 0]

但我敢肯定这不是正确的方法。有什么建议吗？

Answer 1

您可以使用 UTF-16LE 数据表示

let password = "password€"
let data = password.data(using: .utf16LittleEndian)!
print(data as NSData)
// <70006100 73007300 77006f00 72006400 ac20>

这已经足以计算 SHA1 摘要(代码来自 How to crypt string to sha1 with Swift? ):

var digest = [UInt8](repeating: 0, count:Int(CC_SHA1_DIGEST_LENGTH))
data.withUnsafeBytes { 
    _ = CC_SHA1($0, CC_LONG(data.count), &digest)
}
let hexEncodedDigest = digest.map { String(format: "%02hhx", $0) }.joined()
print(hexEncodedDigest)
// 177f0d080dfe533e102dd67d6321204813cf1b0c

但是如果你需要它作为一个字节数组那么

let bytesArray = data.map { $0 }
print(bytesArray)
// [112, 0, 97, 0, 115, 0, 115, 0, 119, 0, 111, 0, 114, 0, 100, 0, 172, 32]

会起作用。

(为了演示，我附加了一个非ASCII字符，€ = U+20AC 变为 172, 32。)

---

如果你想知道如何将 [UInt16] 数组转换为[UInt8] 数组，这是你可以通过一些指针杂耍来实现的(只有一个副本):

let utf16array = Array("password€".utf16)
print(utf16array)
// [112, 97, 115, 115, 119, 111, 114, 100, 8364]

let bytes = Array(utf16array.withUnsafeBufferPointer {
    $0.baseAddress!.withMemoryRebound(to: UInt8.self, capacity: 2 * utf16array.count) {
        UnsafeBufferPointer(start: $0, count: 2 * utf16array.count)
    }
})
print(bytes)
// [112, 0, 97, 0, 115, 0, 115, 0, 119, 0, 111, 0, 114, 0, 100, 0, 172, 32]